Intuition behind showing that $\{x ∈ \mathbb{R} $ $: f(x) > 0\}$ is countable for $\sum f(x)< \infty$

Question

Problem

In a book on Measure Theory, I encountered the following problem:

Let $f : \mathbb{R} → [0,+∞]$ and suppose that $\sum _{x \in \mathbb{R}} f(x) < ∞$.

Show that the set $\{x ∈ \mathbb{R} : f(x) > 0\}$ is countable.

The solution is provided, although I find the general direction it takes to be quite unintuitive and unclear in parts too. The beginning of the solution is written below alongside my question below it.

Comments

Let $M := \sum _{x∈R} f(x) < ∞$. We claim the set {$x ∈ R : f(x) > \frac{M}{k} $} has at most k elements, hence it is finite.

What is the intuition behind this? When attempting this type of problem, is there a way in which I could have been motivated to make the above conjecture?

The proof itself makes sense when I look through the rest of it, but I find it unlikely that I would have stumbled upon the answer myself from this. Is this just a trick or there is something to learn from this example that I can take forward when considering similar problems of this nature?

Terminology

To clarify for those unsure about the definition used for the sum over the real numbers, we use the following definition:

$$ \sum _{x \in \mathbb{R}} f(x) = \sup _{\{\text{finite} \space A \subseteq \mathbb{R}\}} \sum_{x \in A}f(x) $$

(as mentioned in the comments by PrincessEev).

Answer 1

From $\sum_{x\in\mathbb R}f(x)<\infty$ it follows directly that for every positive integer $n$ the set: $$A_n:=\left\{x\in\mathbb R\mid f(x)\geq\frac1n\right\}$$ is finite.

Then based on: $$\{x\in\mathbb R\mid f(x)>0\}=\bigcup_{n=1}^{\infty}A_n$$we find directly that the set $\{x\in\mathbb R\mid f(x)>0\}$ is a union of finite sets hence is countable.

Answer 2

The motivation is essentially that, since we know the range of the function and hence the summation is positive, we can break up the range's values. Several different ways could probably apply, but ultimately the idea here is that

• we want a countable collection of "breaking points".
• we want those "breaking points" to have upper bound $M$ and lower bound $0$; these have to be the bounds, since if $f(x)>M$ for any single $x$, then the sum exceeds $M$, and we know that $f(x) \ge 0$ for all $x$ by definition.
• we want those "breaking points" to be arranged in such a way that the union of the sets they generate is $\{x \mid f(x) > 0\}$.

The "breaking points" in this scenario are the $M/k$'s; notice that this is decreasing in $k$ to $0$, so all three conditions are met. In particular,

$$\{x \mid f(x) > 0\} = \bigcup_{k \in \mathbb{N}} \left\{ x \, \middle| \, f(x) > \frac M k \right\}$$

This construction in particular works nicely for us, because the conclusion we reach (that each set in the union has at most $k$ elements) is almost obvious: if more than $k$ elements were in the set, then the sum would exceed $M$ on just those members of the set! Then it's as simple as the realization that the union of finite sets is at-most-countable.

---

The ultimate trick here is just a matter of partitioning up the range of the function, a common theme in analysis and in particular measure theory. Perhaps even after seeing it, it wouldn't be your first approach, but it becomes more natural the more you see this type of technique applied.

Answer 3

Suppose $\ X:= \{x \in \mathbb{R} : f(x) > 0\}\ $ is uncountable.$\qquad (1)$

Define $\ X_0:=\{x \in \mathbb{R} : f(x) \geq 1\},\ $ and for each $\ n\in\mathbb{N},\ $ define

$\ X_n:= \{x \in \mathbb{R} : \frac{1}{n+1}\leq f(x) < \frac{1}{n}\}.\ $

Suppose further that $\ \vert X_n\vert\ $ is countably infinite for each $\ n\in\mathbb{N}\cup\{0\}.\qquad (2)$

Then $\ X = \displaystyle\bigcup_{n\in\mathbb{N}\cup\{0\}} X_n\ $ would be the countable union of countably many sets, which is countable.

However, since we started assuming $(1),\ X\ $ is uncountable. We have contradicted assumption $(2),$ and so under assumption $(1),\ \exists\ k\in\mathbb{N}\ $ such that $\ \vert X_k\vert\ $ is uncountable.

The result follows because, under assumption $(1),\ \displaystyle\sum_{x\in\mathbb{R}} f(x) \geq \sum_{j\in\mathbb{N}} \frac{1}{k+1} = \infty.$

It would have sufficed to show that $\ \vert X_j\vert\ $ is countably infinite for some $\ j\in\mathbb{N},\ $ but I've done one better.